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Author:  Laurent Buse [ Thu Aug 11, 2005 5:31 pm ]
Post subject:  linear algebra

'tDear Singular users,

'tIs it possible to solve linear systems with
singular ? To be more precise, if A is a given nxm
matrix (of number) and B another given nxk matrix
(of number), is it possible to compute X such that
AX=B ?
'tAlso, what is the more efficient way to compute
the kernel of a matrix of number ? syz(A) ?

Best regards,
Laurent Buse.


email: lbuse@unice.fr
Posted in old Singular Forum on: 2002-08-06 15:31:42+02

Author:  lossen [ Thu Sep 22, 2005 7:43 pm ]
Post subject: 

Dear Laurent,

of course, it is possible to solve systems of linear
equations with SINGULAR. Assume A=(a_{ij}) is a matrix and b
a vector, both with entries in a field F (or, if you
like, of type number). Then you may solve the system
Ax=b in the following manner:

- create the ideal

I = < sum_j (a_{1j}*x_j-b_1) , ... , sum_j (a_{rj}*x_j-b_r) >

and compute a reduced standard basis G of I with respect
to a global monomial ordering (e.g. dp).

- the system Ax=b is solvable over F if and only if
G is not {1}, and then the solutions can be read from G.

You can also solve systems with parameters (i.e., for
different right-hand sides at a time):

Example: Solve the system of linear equations in
x,y,z,u,

3x + y + z - u = a
13x + 8y + 6z - 7u = b
14x +10y + 6z - 7u = c
7x + 4y + 3z - 3u = d

with parameters a,b,c,d. In SINGULAR:

ring R = (0,a,b,c,d),(x,y,z,u),(c,dp);
ideal E = 3x + y + z - u - a,
13x + 8y + 6z - 7u - b,
14x + 10y + 6z - 7u - c,
7x + 4y + 3z - 3u - d;
option(redSB);
simplify(std(E),1); //compute reduced SB

//-> _[1]=u+(6/5a+4/5b+1/5c-12/5d)
//-> _[2]=z+(16/5a-1/5b+6/5c-17/5d)
//-> _[3]=y+(3/5a+2/5b-2/5c-1/5d)
//-> _[4]=x+(-6/5a+1/5b-1/5c+2/5d)

Hence, the (unique) solution is:
x = 1/5 * (6a-b+c-2d),
y = 1/5 * (-3a-2b+2c+d),
z = 1/5 * (-16a+b-6c+17d),
u = 1/5 * (-6a-4b-c+12d).

This and more information about solving you can find in
the new Springer textbook "A SINGULAR Introduction to
Commutative Algebra" (by G.-M. Greuel and G. Pfister).

Concerning your second question: in SINGULAR 'syz(A)' is
the most efficient way to compute the kernel of A.

Best regards, Christoph Lossen.

email: lossen@mathematik.uni-kl.de
Posted in old Singular Forum on: 2002-08-16 13:04:51+02

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